Problem: Simplify and expand the following expression: $ \dfrac{3}{t + 2}- \dfrac{2}{2t + 4}+ \dfrac{3t}{t^2 + 4t + 4} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the second term: $ \dfrac{2}{2t + 4} = \dfrac{2}{2(t + 2)}$ We can factor the quadratic in the third term: $ \dfrac{3t}{t^2 + 4t + 4} = \dfrac{3t}{(t + 2)(t + 2)}$ Now we have: $ \dfrac{3}{t + 2}- \dfrac{2}{2(t + 2)}+ \dfrac{3t}{(t + 2)(t + 2)} $ The least common multiple of the denominators is: $ (t + 2)(t + 2)$ In order to get the first term over $(t + 2)(t + 2)$ , multiply by $\dfrac{2(t + 2)}{2(t + 2)}$ $ \dfrac{3}{t + 2} \times \dfrac{2(t + 2)}{2(t + 2)} = \dfrac{6(t + 2)}{(t + 2)(t + 2)} $ In order to get the second term over $(t + 2)(t + 2)$ , multiply by $\dfrac{t + 2}{t + 2}$ $ \dfrac{2}{2(t + 2)} \times \dfrac{t + 2}{t + 2} = \dfrac{2(t + 2)}{(t + 2)(t + 2)} $ In order to get the third term over $(t + 2)(t + 2)$ , multiply by $\dfrac{2}{2}$ $ \dfrac{3t}{(t + 2)(t + 2)} \times \dfrac{2}{2} = \dfrac{6t}{(t + 2)(t + 2)} $ Now we have: $ \dfrac{6(t + 2)}{(t + 2)(t + 2)} - \dfrac{2(t + 2)}{(t + 2)(t + 2)} + \dfrac{6t}{(t + 2)(t + 2)} $ $ = \dfrac{ 6(t + 2) - 2(t + 2) + 6t} {(t + 2)(t + 2)} $ Expand: $ = \dfrac{6t + 12 - 2t - 4 + 6t}{2t^2 + 8t + 8} $ $ = \dfrac{10t + 8}{2t^2 + 8t + 8}$ Simplify: $ = \dfrac{5t + 4}{t^2 + 4t + 4}$